Problem: The lifespans of seals in a particular zoo are normally distributed. The average seal lives $15.9$ years; the standard deviation is $2.8$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a seal living longer than $10.3$ years.
Answer: $15.9$ $13.1$ $18.7$ $10.3$ $21.5$ $7.5$ $24.3$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $15.9$ years. We know the standard deviation is $2.8$ years, so one standard deviation below the mean is $13.1$ years and one standard deviation above the mean is $18.7$ years. Two standard deviations below the mean is $10.3$ years and two standard deviations above the mean is $21.5$ years. Three standard deviations below the mean is $7.5$ years and three standard deviations above the mean is $24.3$ years. We are interested in the probability of a seal living longer than $10.3$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the seals will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the seals will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $10.3$ years and the other half $({2.5\%})$ will live longer than $21.5$ years. The probability of a particular seal living longer than $10.3$ years is ${95\%} + {2.5\%}$, or $97.5\%$.